2^x+32/2^x=12

Solution for 2^x+32/2^x=12 equation:

2^x+32/2^x=12

We move all terms to the left:

2^x+32/2^x-(12)=0


Domain of the equation: 2^x!=0

x!=0/1

x!=0

x∈R


We multiply all the terms by the denominator


2^x*2^x-12*2^x+32=0

Wy multiply elements

4x^2-24x+32=0

a = 4; b = -24; c = +32;
Δ = b2-4ac
Δ = -242-4·4·32
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8}{2*4}=\frac{16}{8}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8}{2*4}=\frac{32}{8}
=4
$